Question

Consider the function void modify (int \& x) \\[ x=10 \\] \\} Show how to call the modify function so that it sets the integer int i to 10.

Short answer

Answer: To call the `modify` function and set the value of an integer variable 'i' to 10, pass the variable 'i' by reference using the '&' symbol when passing the argument, like this: `modify(i)`.

Step-by-step solution

Declare an integer variable 'i'

Declare an integer variable 'i' which will be the input to the modify function. ``` int i; ```

Call the 'modify' function with 'i' as a reference

Pass the integer variable 'i' by reference to the `modify` function using the '&' symbol. This will allow the function to directly modify the variable's value. ``` modify(i); ```

Verify the value of 'i' after calling the function

After calling the 'modify' function with 'i' passed by reference, the value of 'i' should now be 10. You can verify this by printing the value of 'i' to the console. ``` cout << "The value of i is: " << i << endl; ``` Now after executing the above steps, the modify function will set the value of the integer 'i' to 10.

Key concepts

C++ Function Calls

Function calls are an essential part of programming in C++. When you call a function, you’re asking the program to execute a set of operations defined within that function. To use a function, you must know its name and the parameters it requires. In C++, the basic syntax for a function call is functionName(parameter1, parameter2, ...);. When you call a function, the values you pass to it are called arguments.

For example, in our exercise, the function named modify was designed to take one argument by reference. To call this function, you write modify(i); where i is the integer variable passed as an argument. Once the function is called, the program execution jumps to the body of modify, performs the operations defined there, and then returns control back to the part of the program where the function was called.

C++ Reference Parameters

Reference parameters in C++ are a way to pass the address of a variable to a function, so any changes made to the parameter inside the function affect the original variable. References are declared using the ampersand (&) symbol.

In the context of our exercise, the function modify is defined as void modify(int &x). Here, x is a reference parameter. This means that when we pass a variable to modify, we're not providing a copy; we're providing a direct link to the original variable i. Any changes made to x within the modify function will directly modify i because x is a reference to i.

This technique is particularly useful when you want a function to alter the value of the passed variable, or when you want to avoid copying large amounts of data for performance reasons.

Modifying Variable Values

Modifying variable values is a common operation in C++, and it can be done directly within a function when using reference parameters. As demonstrated in the exercise, the modify function takes a reference to an integer and sets its value to 10 with the statement x=10;.

The implication of this action is that the original variable passed into the function, in this case, i, will have its value changed without the need for a return value or additional assignments. Modifying variables in this way ensures that any changes are reflected in the caller’s scope, which is powerful for functions that need to update multiple values or large data structures.

It’s important to note that when using references to modify variables, you should be aware of the side effects this may have on your program. By allowing functions to change the values of variables passed to them, you can have more compact and efficient code, but it can also introduce bugs if not managed carefully.